Do uncountable spaces admit Markov strategies in Rothberger-style games?

Question

Consider the selection game $G_1(\mathcal C,\mathcal C)$ where $\mathcal C$ is associated with a class of open covers. Considering various possibilities for $\mathcal C$, do uncountable spaces admit winning Markov (relying on only the round number and the most recent move of the opponent) strategies for the second player?

(The answer will be modulo separation axioms: of course, any indiscrete space of arbitrary cardinality only allows the trivial open cover, and thus the second player has a trivial winning Markov strategy.)

Answer 1

The case when $\mathcal C = \mathcal K$

If $\mathcal C = \mathcal K$, the set of compact subsets, then the reals are an example of an uncountable space where the second player has a winning Markov strategy in the game $\mathsf G_1(\mathcal C, \mathcal C)$. In fact, any uncountable hemicompact space is such an example.

The case when $\mathcal C = \Omega$

By the duality results of Clontz' Dual Selection Games, the game $\mathsf G_1(\Omega,\Omega)$ is dual to the game $\mathsf G_1(\mathcal F, \neg \Omega)$ where $\Omega$ is the collection of $\omega$-covers of a space (all nontrivial open covers that capture each finite set) and $\mathsf G_1(\mathcal F,\neg \Omega)$ is the game where the first player chooses finite sets $F_n$ and the second player chooses open sets $U_n$ so that $F_n \subseteq U_n$. The first player wins $\mathsf G_1(\mathcal F, \neg \Omega)$ if the second player produces an $\omega$-cover. This is similar to the finite-open game.

As a similar example as in Steven Clontz' answer, the right-ordered reals is an example of an uncountable $T_0$ space in which P1 has a predetermined winning strategy in $\mathsf G_1(\mathcal F, \neg \Omega)$: P1 can simply select $\{-n\}$ in the $n^{\mathrm{th}}$ inning. Hence, the second player has a winning Markov strategy in $\mathsf G_1(\Omega,\Omega)$.

However, like before, being $T_1$ guarantees countability.

Assume $X$ is $T_1$ and that P1 has a winning predetermined strategy in $\mathsf G_1(\mathcal F, \neg\Omega)$ (equivalently, that P2 has a winning Markov strategy in $\mathsf G_1(\Omega,\Omega)$). Then we can let $F_n$ be the choice P1 makes in the $n^{\mathrm{th}}$ inning according to their predetermined winning strategy. We will show that $X = \bigcup_{n\in\omega} F_n$. If there were some $x \not\in \bigcup_{n\in\omega} F_n$, then $X \setminus \{x\}$ would be a valid attack against the first player's strategy which contradicts the fact that it is winning. So $X$ is countable.

Answer 2

P2 has a winning Markov strategy in the Rothberger game $G_1(\mathcal O,\mathcal O)$ (where $\mathcal O$ collects all open covers of the space) if and only if the space is "topologically countable". That is, there exists a countable set $\{x_n:n<\omega\}$ such that for each point $x\in X$, there is $n<\omega$ such that every neighborhood of $x_n$ also contains $x$.

It will be easier to consider the game $G_1(\mathcal P,\neg\mathcal O)$, where P1 chooses points and P2 chooses open neighborhoods, and where P1 (not P2) wins if P2's selections form an open cover. This game is dual to $G_1(\mathcal O,\mathcal O)$, which means a winning Markov strategy for P2 in $G_1(\mathcal O,\mathcal O)$ exists if and only if a winning predetermined (relying on only the round number) strategy for in $G_1(\mathcal P,\neg\mathcal O)$ exists.

So if the space is topologically countable, P1's winning perdetermined strategy is simple: choose $x_n$ during round $n$. Then any selections by P2 must form an open cover by the definition of topologically countable. Likewise given a winning predetermined strategy, the points chosen by the strategy witness topological countability.

So then for $\mathcal C=\mathcal O$, the question reduces to when topological countability implies actual countability. $T_0$ is insufficent: consider the non-negative reals with the rightward topology (a set is open if and only if for each point in the set, all greater real numbers are also in the set). Then $\{0\}$ witnesses topological countability, but the space is not countable.

Assuming $T_1$, we have the desired result. Given a topologically countable space witnessed by $\{x_n:n<\omega\}$, let $x\in X$. There exists $n<\omega$ where every open neighborhood of $x_n$ contains $x$. But $X\setminus\{x\}$ is open, so it cannot contain $x_n$. Thus $x=x_n$, and $X=\{x_n:n<\omega\}$.

Thus for $\mathcal C=\mathcal O$, the answer is "yes" for a $T_0$ space, and "no" for all $T_1$ spaces.

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EDIT: Let's extend this result to show Markov strategies for P2 in $G_1(\Omega,\Omega)$ also characterize topological countability (without any separation axioms assumed). Again, it's easier to consider the dual game $G_1(\mathcal F,\neg\Omega)$ and P1's predetermined strategies. In this game, P1 chooses finite sets, P2 chooses open neighborhoods, and P1 wins if P2 constructs an $\omega$-cover (every finite set is contained in some open set).

Given topological countability, let $\{F_n:n<\omega\}$ enumerate the finite subsets of $\{x_n:n<\omega\}$. Then this is a winning predetermined strategy for P1: given a finite set, each point has an associated $x_n$, and these finitely-many $x_n$ were handled during some round.

Now let $\{F_n:n<\omega\}$ be a winning predetermined strategy for $P1$. Suppose there was $x\in X$ such that for each $n<\omega$ there was an open set $U_n\supseteq F_n$ with $x\not\in U_n$. Then this is exactly a counterplay by P2 that defeats the predetermined strategy: $\{x\}$ is a finite set not contained in any open set chosen by P2.